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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?

Resources

The Resources for this project are background ideas that you will need to understand and make progress on the CrowdMath problems. We've chosen resources that are directly relevant to the project: the problems are defined explicitly in terms of ideas that you will find in our resources. All of the problems we will pose are thematically linked and all of the resources we post will be as well. We will be releasing the resources roughly in the order that you should be reading them.

Each resource also has exercises to help clarify the key ideas and give practice with them. You can discuss the exercises by clicking either the "View Discussions" or "Start New Topic" button. You can discuss any resource by clicking the V or t buttons on the resources.

V t Resource 0: Preliminaries
Synopsis: In this first resource, we introduce the problem we are interested in as well as the primary algebraic structure we will be dealing with in the scope of this new research project.

The Goldbach Conjecture

The Goldbach conjecture, introduced by Goldbach in a 1742 letter to Euler, remains one of the most ancient and renowned unsolved problems in mathematics. Modernly stated, the conjecture asserts that every even integer greater than $2$ can be expressed as the sum of two prime numbers.

Analogues of the Statements of Goldbach Conjecture

Since the 1960s, several authors have explored analogues of the Goldbach conjecture for various classes of polynomial rings. In 1965, Hayes demonstrated that polynomials $f \in \mathbb{Z}[x]$ with $\deg(f) > 1$ can be represented as the sum of two irreducibles. This work was later extended by Saidak, Kozek, and Pollack. Additionally, variations of the Goldbach conjecture for polynomials over finite fields have been investigated by Effinger and Hayes, Bender, Car and Gallardo, among others. More recently, Paran examined a version of the Goldbach conjecture in the ring of formal power series over the integers.

Semirings and Semidomains

A commutative semiring $S$ is a nonempty set endowed with two binary operations denoted by `$+$' and `$\cdot$' and called addition and multiplication, respectively, such that the following conditions hold:

(1) $(S,+)$ is a monoid with its identity element denoted by $0$;

(2) $(S, \cdot)$ is a commutative semigroup with an identity element denoted by $1$ with $1 \neq 0$;

(3) $b \cdot (c+d)= b \cdot c + b \cdot d$ for all $b, c, d \in S$.

Let $S$ be a commutative semiring. Since the operation of addition in $S$ is cancellative, the distributive law ensures that $0 \cdot b = 0$ for all $b \in S$. Throughout our CrowdMath project, for any $b,c \in S$, we write $b c$ instead of $b \cdot c$ when there seems to be no risk of ambiguity. In the typical definition of a `semiring' $S$, one does not assume that the semigroup $(S,+)$ is cancellative. However, we do so here because our semirings of interest have cancellative additive structures. On the other hand, it is worth observing that a more general notion of a `semiring' $S$ does not assume that the semigroup $(S, \cdot)$ is commutative. However, once again, this more general type of algebraic objects are not of interest in the scope of this paper. Accordingly, from now on we will use the single term semiring to refer to a commutative semiring, tacitly assuming the commutativity of both operations. A subset $S'$ of $S$ is a subsemiring provided that $(S',+)$ is a submonoid of $(S,+)$ that is closed under multiplication and contains $1$. Observe that every subsemiring of $S$ is a semiring. As an exercise, you will show that the set $S[x]$ (resp., $S[x^{\pm 1}]$) consisting of all polynomials (resp., Laurent polynomials) with coefficients in $S$ is also a semiring under the standard addition and multiplication of polynomials (resp., Laurent polynomials). We call $S[x]$ (resp., $S[x^{\pm 1}]$) the semiring of polynomials (resp., semiring of Laurent polynomials) over $S$.

Definition. We say that a semiring $S$ is a semidomain provided that $S$ is a subsemiring of an integral domain.

Let $S$ be a semidomain. Then $(S \setminus \{0\}, \cdot)$ is a monoid, which we denote by $S^*$ and call the multiplicative monoid of $S$. In the next example, we show a semiring that is not a semidomain but still its subset of nonzero elements is a multiplicative monoid. In order to reuse notation from ring theory, we refer to the units of the monoid $(S,+)$ as invertible elements of the semidomain $S$, so that we can refer to the units of the multiplicative monoid $S^*$ simply as units of $S$, avoiding any risk of ambiguity. Also, following standard notation from ring theory, we let $S^\times$ denote the group of units of $S$, letting $\mathcal{U}(S)$ refer to the additive group of invertible elements of $S$. In addition, we write $\mathcal{A}(S)$ instead of $\mathcal{A}(S^*)$ for the set of atoms of the multiplicative monoid $S^*$. Finally, for any $b,c \in S$ such that $b$ divides $c$ in $S^*$, we write $b \mid_S c$ instead of $b \mid_{S^*} c$, and the term (greatest, maximal) common divisor in the context of semidomains are to be understood in the multiplicative monoid $S^*$.

For the next example, we need the following lemma.

Lemma. For a semiring $S$, the following conditions are equivalent.

(a) $S$ is a semidomain.

(b) The multiplication of $S$ extends to the Grothendieck group $\mathcal{G}(S)$ of $(S,+)$ turning $\mathcal{G}(S)$ into an integral domain.


Here is an educative example.

Example. Notice that the set $S := \{(0,0)\} \cup (\mathbb{N} \times \mathbb{N})$ is a monoid with the usual component-wise addition, and it is closed under the usual component-wise multiplication with multiplication identity $(1,1)$. Hence $S$ is a semiring. Observe, on the other hand, that any extension of the multiplication of $S$ to $\mathcal{G}(S) = \mathbb{Z} \times \mathbb{Z}$ making the latter a commutative ring must respect the identity $(1,0) (0,1) = (0,0)$ and, therefore, will not turn $\mathcal{G}(S)$ into an integral domain. Hence it follows from the previous lemma that $S$ is not a semidomain.
V t Exercise 0.1
Prove that for a semiring $S$, the following conditions are equivalent.

(a) $S$ is a semidomain.

(b) The multiplication of $S$ extends to the Grothendieck group $\mathcal{G}(S)$ of $(S,+)$ turning $\mathcal{G}(S)$ into an integral domain.
Discussion Exercise:
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V t Exercise 0.2
Prove that, for each $n \in \mathbb{N}$, there exists a polynomial in the semidomain $\mathbb{N}_0[x]$ that cannot be expressed as the sum of $n$ irreducibles.
Discussion Exercise:
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V t Exercise 0.3
Consider the "almost semidomain" $S := (\mathbb{N}_0 \setminus \{1\})[x^{\pm}]$ ($S$ is not a semidomain because it does not contain $1$). Prove that there exist infinitely many Laurent polynomials in $S$ that cannot be written as the sum of two irreducibles.
Discussion Exercise:
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V t Exercise 0.4
Let $S$ be a semidomain. Prove the following statements.

(1) The polynomial extension $S[x]$ is also a semidomain.

(2) The Laurent polynomial extension $S[x^{\pm 1}]$ is a semidomain.

(3) The power series extension $S[[x]]$ is a semidomain.

(4) If the additive monoid $(S,+)$ is reduced, then $(S[x]^*, \cdot)$ is a divisor-closed submonoid of $(S[[x]]^*, \cdot)$.
Discussion Exercise:
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V t Resource 1: A First Insight into Goldbach's Statements
Semidomains VS Integral Domains

Liao and Polo initiated the study of analogues of the Goldbach conjecture for polynomial semirings. Specifically, they showed that every $f \in \mathbb{N}_0[x^{\pm 1}]$ that is not a monomial can be written as the sum of at most two irreducibles provided that $f(1) > 3$. It is worth mentioning that the proof uses some facts about the distribution of prime numbers in $\mathbb{Z}$. Note that this theorem is closely connected to the Goldbach conjecture. This result is about partitioning a fixed set of units into two subsets, where each subset represents an irreducible Laurent polynomial with positive integer coefficients, just as the Goldbach conjecture concerns the partitioning of a fixed set of units into two subsets, where each subset represents a positive prime. Moreover, if the Goldbach conjecture were true then, using that $f \in \mathbb{N}_0[x^{\pm 1}]$ is irreducible when $f(1)$ is a prime number, one could prove that Laurent polynomials $f \in \mathbb{N}_0[x^{\pm 1}]$ for which $f(1)$ is an even number greater than $2$ can be written as the sum of two irreducibles.

As noted in Resource $0$, Hayes was the first to explore analogues of the Goldbach conjecture in the context of polynomial rings. In fact, he showed that polynomials $f \in \mathbb{Z}[x]$ with $\deg(f) > 1$ can be represented as the sum of two irreducibles. But, how does this statement relate to the Goldbach conjecture? For starters, in the context of Hayes' statement we are allowed to subtract. This somehow changes the nature of the Goldbach conjecture as elements might have various representations as sum of units. For instance, the polynomial $x^3 + 1$ can be also represented as $x^3 + 3 - 2$. Observe that in the context of the original Goldbach conjecture $11 - 3$ is not a valid Goldbach decomposition of the number $8$. On the other hand, assuming the Goldbach conjecture to be true does not provide much additional insight into Hayes's statement.


Polynomial Semidomains vs Laurent Polynomial Semidomains

As part of the exercises of Resource $0$, you proved that $\mathbb{N}_0[x]$ and $\mathbb{N}_0[x^{\pm 1}]$ are semidomains. However, observe that no analogue of the Goldbach conjecture holds for $\mathbb{N}_0[x]$. Indeed, since $x$ is an additive atom that is not a multiplicative unit, then the polynomial $x^4 + x^3 + x^2 + x$ cannot be expressed as the sum of two irreducibles (see Exercise $2$). In contrast, in the context of $\mathbb{N}_0[x^{\pm 1}]$ we can write
\[
	x^4 + x^3 + x^2 + x = (x^4 + x^3) + (x^2 + x),
\]where both expressions inside parentheses are irreducibles in $\mathbb{N}_0[x^{\pm 1}]$. The key difference is that now the element $x$ is a multiplicative unit. For this reason, we focus on Laurent polynomial semidomains rather than polynomial semidomains.

One of the difficulties one encounters in establishing analogues of the Goldbach conjecture for polynomial semirings is finding suitable irreducibility criteria. Observe that, even though a semidomain $S$ can be nicely embedded into an integral domain $\mathcal{G}(S)$, the irreducibles of $S$ does not coincide with the irreducibles of $\mathcal{G}(S)$. For instance, the polynomial $f = x^3 + 1$ is irreducible in $\mathbb{N}_0[x^{\pm 1}]$ but, clearly, $f$ can be decomposed as $(x + 1)(x^2 - x + 1)$ when considered as a polynomial with coefficients in $\mathbb{Z}$. Additionally, note that the units of $S$ and $\mathcal{G}(S)$ may differ. As a matter of fact, the series $\sum_{i = 0}^{\infty} x^i$ is a unit in $\mathbb{Z}[[x]]$ but it is not a unit in $\mathbb{N}_0[[x]]$. We can then conclude that the factorization properties of $S$ and $\mathcal{G}(S)$ are different in general.
V t Exercise 1.1
Prove that every polynomial $f \in \mathbb{Z}[x]$ with $\deg(f) > 1$ can be represented as the sum of two irreducibles.
Discussion Exercise:
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V t Exercise 1.2
Let $S$ be an additively reduced semidomain containing an additive atom that is not a multiplicative unit. Prove that no analogue of the Goldbach conjecture hold for $S[x,x^{-1}]$.
Discussion Exercise:
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V t Exercise 1.3
Provide two irreducibility criteria for polynomials in $\mathbb{N}_0[x]$.
Discussion Exercise:
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V t Exercise 1.4
Find an almost semidomain (i.e., a semidomain possibly missing a multiplicative identity) $S$ for which $S[x^{\pm 1}]$ does not satisfy an analogue of the Goldbach conjecture.
Discussion Exercise:
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V t Localization and the Goldbach property
Synopsis: Some of the semidomains we have been studied consist of positive rational numbers. For instance, the semidomain $\mathbb{N}_0\Big[\frac{1}{2}\Big]$ consists of all dyadic nonnegative rationals. A formal algebraic way of introducing denominators for a given semidomain (or, more generally, semiring) is through localizations. In this resource we will introduce the notion of localization in the setting of semidomains, and in the corresponding open problem section, we propose some problems about Goldbach analogues in connection with localization.

Localization of Semidomains

Let $\mathfrak{S}$ be a semidomain, and let $S$ be a multiplicative subset of $\mathfrak{S}$ (i.e., a submonoid of $\mathfrak{S}^*$). Observe that $S$ is also a multiplicative subset of $\mathcal{G}(\mathfrak{S})$. Now set $R := (\mathfrak{S} \times S)/\sim$, where $\sim$ is the equivalence relation on $\mathfrak{S} \times S$ defined as follows: $(s,d) \sim (s',d')$ if and only if $sd' = ds'$ (you will formally prove this in Exercise 2.1). We let $\frac{s}{d}$ denote the equivalence class of $(s,d)$. Now define the following binary operations $R$:

(i) $\frac{s}{d} \cdot \frac{s'}{d'} = \frac{ss'}{dd'}$

(ii) $\frac{s}{d} + \frac{s'}{d'} = \frac{sd' + ds'}{dd'}$.

It is routine to verify that these operations are well defined and that $(R,+,\cdot)$ is a semiring (we propose this as Exercise 2.2(a)). In fact, it is possible to prove that $(R,+,\cdot)$ is a semidomain (this is Exercise 2.2(b)). The semidomain $R$ is called the localization of $\mathfrak{S}$ at the multiplicative set $S$, and it is denoted by $S^{-1}\mathfrak{S}$.

Let us take a look at a couple of simple examples of localization in the setting of semidomains. this construction seems very abstract!

Example 1: Consider the semidomain $\mathbb{N}_0$ consisting of all nonnegative integers under the standard addition and. multiplication, and let $S$ denote the set of positive integers, which clearly a multiplicative subset of $\mathbb{N}_0$. What is $S^{-1}\mathbb{N}_0$? Well, according to the previous construction, the elements of this new semidomain are fractions $\frac{n}{d}$ where $n \in \mathbb{N}_0$ and $d \in \mathbb{N}$, and we say that $\frac{n}{d} = \frac{n'}{d'}$ provided that $nd' = dn'$. This set with operations defined as in (i) and (ii) above looks a lot like the semidomains consisting of all nonnegative rational numbers. According to Exercise 2.3, this semidomain is $\mathbb{Q}_{\ge}$.


Example 2: Let us start once again with the semidomain $\mathbb{N}_0$, but now fix a prime $p$ and consider the subset $S := \big\{p^k : k \in \mathbb{N}_0 \big\}$. Clearly, $S$ is a multiplicative subset of the semidomain $\mathbb{N}_0$, and it is not hard to see that the elements of $S^{-1}\mathbb{N}_0$ are fractions $\frac{n}{d}$ with $n$ being a nonnegative integer and $d$ being a nonnegative power of $p$. As it turns out, this semidomain is isomorphic to $\mathbb{N}_0\Big[\frac{1}{p}\Big]$, we it is stated in Exercise 2.4.


Observe that the elements of $S^{-1}\mathfrak{S}$ are fractions of the form $\frac{s}{d}$, with $s \in \mathfrak{S}$ and $d \in S$. Moreover, it is not hard to show that all elements of the form $\frac{d}{s}$ with $d,s \in S$ are units in $S^{-1}\mathfrak{S}$. As a matter of fact, this is the ``simplest extension" of the semidomain $\mathfrak{S}$ for which all elements of $S$ become units (this is the order of Exercise 2.5).
V t Exercise 2.1
Prove that $\sim$ is an equivalence relation.
Discussion Exercise:
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V t Exercise 2.2
Solution:
(a) Verify that the two binary operations that we defined and denote by "+" and "$\cdot$" in the construction of localization are well defined and turn $R$ into a semiring.

(b) Show that $(R,+, \cdot)$ is actually a semidomain.
Discussion Exercise:
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V t Exercise 2.3
Let $S$ and $T$ be two semirings, and let $\rho\colon S \to T$ be a map. We say that $\rho$ is a semiring homomorphism if for all $s_1, s_2 \in S$, the following conditions hold:

(i) $\rho(s_1 + s_2) = \rho(s_1) + \rho(s_2)$;

(ii) $\rho(s_1 s_2) = \rho(s_1)\rho(s_2)$.

If, additionally, the map $\rho$ is a bijection (i.e., simultaneously injective and surjective), then we say that $\rho$ is a semiring isomorphism and also that the semirings $S$ and $T$ are isomporphic.

(a) Prove that being isomorphic determines an equivalence relation in the class consisting of all semirings.

(b) Prove that the semidomain defined in Example 1 is isomorphic to the semiring $\mathbb{Q}$.
Discussion Exercise:
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V t Exercise 2.4
Prove that the semidomain defined in Example 2 is isomorphic to $\mathbb{N}_0\Big[\frac{1}{p}\Big]$.
Discussion Exercise:
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V t Exercise 2.5
Let $\mathfrak{S}$ be a semidomain, and let $S$ be a multiplicative subset of $\mathfrak{S}$. Prove that the following statements hold.

(a) The map $\pi \colon \mathfrak{S} \to S^{-1}\mathfrak{S}$ given by $\pi(t) = \frac{t}{1}$ is an injective semiring homomorphism satisfying that $\pi(s)$ is a unit in $S^{-1}\mathfrak{S}$ for all $s \in S$.

(b) If $\theta \colon \mathfrak{S} \to \mathfrak{T}$ is a semiring homomorphism such that $\theta(s)$ is a unit in $\mathfrak{T}$ for every $s \in S$, then there exists a unique semiring homomorphism $\phi \colon S^{-1}\mathfrak{S} \to \mathfrak{T}$ such that $\phi \circ \pi = \theta$.
Discussion Exercise:
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